Reading the voltage across the resistor and I get 1.5A, and then slowly after about 20 seconds climbs to 1.67A... however after 30 seconds the test load is getting way too hot, the diodes and resistor can't be touched. So I turn it off.

Should the driver be going straight to 1.8A? Is the driver bad?
Are my batteries not able to provide enough current? I have two good sanyo's I bought, however their contact points are shallow so I'd need to rig them up somehow else..

The IDEA of the test load is to use a cheaper diode with the same electrical properties. Since it's hard to find another diode that drops that many volts, we use a bunch of regular, cheap diodes in series, and add up their Vf (@ whatever current you will be running them at) to as close an approximation of the Vf of the laser diode at that current. The resistor is put in place simply to get a reading of teh current without affecting the circuit too much. We use 1 Ohm resistors because everybody is dumb and can't do multiplication or division.

It all depends on the current that goes throught the diodes. Larger currents will result in larger voltage drop. It's not three for 445nm or four for 445nm...

It can be 5 or 6 for 445nm running at 200mA for example. The more current the more voltage drop across each diode. Do some experimenting reading the voltage with different currents and diodes. It's easy stuff

I think I'm still a little thick on something, so if someone could please correct me.

The forward voltage (Vf) is the voltage a component uses to operate. So my 445 diode has a different Vf for a different current. So lets say I have an A140(more on this later), I would need to find from a chart what the Vf is for a given current setting. This will give me the Vf I want to "simulate". Lets say its 5V. Jufrans' 3A selectable test load uses 1N5404 diodes which have a drop of 1.2V. So to get as close as possible, I want to set it to pin 4.
Was all that right? I hope? Does the source voltage play a roll in these calculations? For example if I'm using a linear driver, I have an 8.4V source, or if I have a buck-boost then source is always what the diode wants...

Just a little more clarification perhaps!

Thanks

p.s- the diode I am working with is the 445 diode that was in survival lasers original laser, in the C6 host... anyone know what diode that is ?

How do you figure 1.2V? It's closer to 0.9V according to the datasheet. Don't forget the current sense resistor drops 1V for every ampere of current through it, so you need 3 diodes. If I assume you're over-driving the piss out of it like everyone else, that's 1.8V + 3*0.9V = 4.5V

The source voltage does not play a role. However, it must be in the operating range of the driver.

You can't call your source 8.4V if it's two lithium ion cells. That's the full voltage, yes, but it falls very quickly to ~7.2V, and if you plan on utilizing the full capacity of the cells, you should design for a lower limit of 5.6V.

At 1.8A the testload will heat up very quickly. At 20-30 seconds it is already blistering hot.
Unless you heatsink the testload and remove the heat buildup quickly, you can't make long testing on it. It is meant for testing if the current is correct. Not for duration testing of the driver.

The voltage drop of your components depends on the current flowing through them. For half amp, you can expect closer to 0.8V for diodes and 0.5V for the resistor. So maybe 7 ideally? 6 is fine. So is 5. It doesn't matter much if it's a linear driver and if it is heat sinked well.

The voltage drop of your components depends on the current flowing through them. For half amp, you can expect closer to 0.8V for diodes and 0.5V for the resistor. So maybe 7 ideally? 6 is fine. So is 5. It doesn't matter much if it's a linear driver and if it is heat sinked well.

This is something I am trying to explain to everyone (that depending on the current different voltage drops occur and you have to use that many diodes depending on the current and voltage you are trying to simulate) So far I can't see if anyone actually understood me

Just like you checked to see what the Vf was at the current you will be running it at, you need to do the same for the diodes in your test load to know how many to use. Remember, they're both diodes.

Some people respond positively to snide questions that simultaneously insult them:
Why would you go through all this work to make sure you know the right Vf at that current, and then just assume a Vf for your test load, unless you are some kind of n00b?

Using the chart at the end we just read off the Vf at a given current setting. Use that Vf.
The resistor voltage drop will just be whatever current we are working at, because its a 1ohm resistor. (0.5A=0.5V.. etc)